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Applied Electromagnetic Waves

by: Karolann Wiegand

Applied Electromagnetic Waves ECE 3317

Marketplace > University of Houston > Electrical Engineering > ECE 3317 > Applied Electromagnetic Waves
Karolann Wiegand
GPA 3.51

David Jackson

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David Jackson
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This 162 page Class Notes was uploaded by Karolann Wiegand on Saturday September 19, 2015. The Class Notes belongs to ECE 3317 at University of Houston taught by David Jackson in Fall. Since its upload, it has received 11 views. For similar materials see /class/208291/ece-3317-university-of-houston in Electrical Engineering at University of Houston.

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Date Created: 09/19/15
ECE 3317 Prof David R Jackson Notes 1 Waveguidmg Structures Chapter 5 A waveguiding structure is one that carries a signal or power from one point to another There are three common types Transmission lines Fiberoptic guides Waveguides Properties Has two conductors running parallel Can propagate a signal at any frequency in theory Becomes lossy at high frequency Can handle low or moderate amounts of power Does not have signal distortion unless there is loss May or may not be immune to interference Does not have E2 or HZ components of the fields TEMZ y RjaLGjaC k2 j j RjaLGjaC J a Lossless kl 0 LC 2 a ya k always real OL 0 Properties Has a single dielectric rod Can propagate a signal at any frequency in theory Can be made very low loss Has minimal signal distortion Ver immune to interference Not suitable for high power Has both E and Hz components ofthe fields Two types of fiberoptic guides 1 Singlemode fiber Carries a single mode as with the mode on a waveguide Requires the fiber diameter to be small relative to a wavelength 2 Multimode fiber Has a fiber diameter that is large relative to a wavelength It operates on the principle of total internal reflection critical angle effect MultiMode Fiber httpenwikipediaorgwikiOptica ber Properties Has a single hollow metal pipe Can propagate a signal only at high frequency a gt 6 The width must be at least one half of a wavelength Has signal distortion even in the lossless case Immune to interference Can handle large amounts of power Has low loss compared with a transmission line Has either E2 or HZ component of the fields TMZor TEZ http en wikiped ia orgwikiWaveg u ideelectromag netism kc constant k a 8 wavenumber of material inside waveguide We can write kc QC 8 wavenumber of material at cutoff frequency 06 a gt we k2 k2 k3 2 real propagation a lt we k2 j k k2 imaginary evanescent decay Statement All six field components of a guided wave can be expressed in terms of the two fundamental field components EZ and HZ quotGuidedwave theoremquot Assumption This is the definition of a guided wave A proof of this statement is given next I Proof illustrated for By I Vx fw5E E 1 aHZaij y jag 6x 62 or Ey 1 6HZszij Jaw 536 Now solve for Hx VXEjw 8E Hx 1 6EZ y jam 8y 82 jaw 6y y Substituting this into the equation for Ey yields the result E aHZ kz 8E2 4sz y jaw 6x jaw 6y y Next multiply by jayjag k2 This gives us 8E2 Z X szy jwya jkz k Ey Solving for By we have The other three components Ex Hx Hy may be found similarly Summary of Fields jau 6 Ex k2 42 Hz jkz 6E2 6 CZ k 6x sz 6 kZ kf ax kZ k E a 2 a B By a y jam 5 Z sz HZ Hx 2 2 2 k kz k kZ x 8H Hy ja8 jkz Z CZ k 6x CZ k 6y To avoid havir lg a completely zero field k k Hehce I Examples of TEMZ waves A wave in a transmission line A plane wave llll ll lllll rl ZQW WW S plane wave IE Im lWave Impedance Property of TEMZ Mode Faraday39s Law V X E jau1j 3 8E Take the x component of both sides E y jauHx ay 82 The field varies as Ey xy Z Ey0 xy 6 ij Hence jkEy jauHx E Therefore we have y a u a 77 Hx k a lug 8 WE jw1j Now take they component of both sides LE E jaH 8x 62 y Hence jkEx jatu Therefore we have Ex 2 a 2 J2 77 H k mJug 8 y Hence Summary I hese two equations may be written as a single vector equationquot E 7sgtlt51 l he electric and magnetic fields of a I EMZ wave are perpendicular to each other and the amplitudes of them are related by 77 Examples plane wave IE Im The fields look like a plane wave in the central region m icrostrip I Ir a waveguide the fields Qahhot be I EMZ I y T DEC bouhdar y proof E I C 3 Assume a EMZ field x g Waveg u iczle B IE d gt 0 property of flux Iihe IQ jE 62739 gt O 4 C cohtr adictioh 5393 A Q3 4E dr 7 72 dS I Z S O Faraday39s law lr I lr Itegr al form C S at S at In a waveguide there are two types of fields TMZ HZ 0 E27 0 TEZ E2 0 H27 0 ECE 3317 Prof David R Jackson Notes 5 Poynting Theorem Adapted from notes by Prof Stuart A Long The Poynting theorem is one on the most important in EM theory It tells us the power flowing in an electromagnetic field John Henry Poynting 18521914 John Henry Poynting was an English physicist He was a professor of physics at Mason Science College now the University of Birmingham from 1880 until his death He was the developer and eponym ofthe Poynting vector which describes the direction and magnitude of electromagnetic energy flow and is used in the Poynting theorem a statement about energy conservation for electric and magnetic fields This work was first published in 1884 He performed a measurement of Newton39s gravitational constant by innovative means during 1893 In 1903 he was the first to realize that the Sun39s radiation can draw in small particles towards it This was later coined the PoyntingRobertson effect In the year 1884 he analyzed the futures exchange prices of commodities using statistical mathematics Wikipedia Subtract and use the following vector identity VxE EVx VEx We then have VEXEFJE a gEa g VEXEFJE a gEa g Next assume that Ohm39s law applies for the electric current 102 gt VEx 0E E a EEa Q at at or VEx 0 E 2 Ha E E62 VEXEFUlElz g gEg g From calculus chain rule we have that at at 26t Hence we have VEx 0El ltHagt e ltE VEx 0EIZ13 85 EE 18 2 6t 6t This may be written as VEx 0E213 lz8 2 at l a EI2 25 or a 1 vltExagt a rEgylarj ggewz Final differential point form of Poynting theorem a 1 1 V Ex W g lzj Volume integral form Integrate both sides over a volume and then apply the divergence theorem 6 1 a 1 V Ex 0E2 g lzj gggl lzj JVEx dV 0 2dV y zjdV lg zjdV Final volume form of Poynting theorem 43Ex S 6 t I V dS I0E2dV V 1 H 2 For a stationary surface 95 S a 1 2 a 1 2 0A 2 Physical interpretation Assume that S is stationary A a 1 a 1 ltjgt gtlt ndS jag2 dV I u 2 dV I g 2 dV S V a V 2 a V 2 Power dissipation as heat Joule39s law Rate of change of stored magnetic energy Rate of change of stored electric energy Righthand side power flowing into the volume of space Hence gt x d5 2 power flowing into the region S Or we can say that Ntin wit it t Define the Poynting vector Analogy gflat 2 power flowing out of the region S g at 2 current flowing out of the region S J current density vector direction of power flow 51 The units of are WmZ l I he power P flowir Ig through the surface S from left to right is PIS d5 S fin W 7 UUJKiv f Assume sinusoidal timeharmonic fields xyzt Exyztgtlt xy ZJ Exy 21 Re Exyzejm x y 21 Re Hxyzejm From our previous discussion notes 2 about time averages we know that lt tgtlt tx tgtReEij De ne the complex Poyhting vector 1 gtxlt We then have that Note the imaginary part of the complex Poyhtihg vector corresponds to the VARS flowing in space Although the Poynting vector can always be used to calculate power flow at low frequency circuit theory can be used and this is usually easier Example 39 39 V ECE 3317 Prof David R Jackson Notes Transmission Lines Frequency Domain Chapter 6 Why is the frequency domain important modulate I Most communication systems use a sinusoidal signal which may be d I A solution in the 39equency domain allows to solve for an arbitrary time varying signal on a lossy line by using the Fourier transform method mo Vte dt Vt i if 1760 emdw AZ C capacitancelength Fm L inductancelength Hm R resistancelength Qm G conductancelength Sm Take the derivative of the first TE with respect to z Substitute in from the second TE GV C K a 822 a V RGV RCLG LC 82V 52 i0 Recall no exact solution in the time domain for the lossy case MLAHWI To convert to the phasor domain we use a gt ja l 52V RGV RCLGa V LC82V 0 522 at 82 82V 2 822 RGV jaRCLGV ja LCV 0 or dZV d 2 RGV jaRC LGV a2LCV Z sz d 2 RGV jaRC LGV a2LCV Z Note that RG jaRC LG a2LC R jaJLG ij Z R jaL series impedancelength Y G ij parallel admittancelength Then we can write d V ZYV d 2 Solutior I Note I her IVCZ We have ah exact solutior t Convention we choose the complex square root to be the principle branch lossy case y is called the propagation constant with units of Wm Choosing the principle branch means that Rey 2 O For a lossless line we consider this as the limit of a lossy line in the limit as the loss tends to zero lossless case Physical interpretation of waves V Z A 9 72 forward traveling wave V Z B 97 forward traveling wave This will be shown shortly Denote y propagation constant 1m a attenuation constant nepersm Note or 0 for a lossless line 8 phase constant radiansmeter Fonvard traveling wave V z fie dz ej z Special case of a lossless line yzjwVLC Compare with 7 06 Hence we realize that Forward traveling wave Denote A Alem Then Vz Alej ZS em 6 132 In the time domain we have Vzt ReV 26W Hence we have I he dietar nce 7x V Z t eiO Z cosat 62 I 8 hapehot of Waveform 2L eez e Iz lle Z is the dietar nce it takes for the waveform to repeat I x wavelehgthr itee If ir n meters The wave repeats Hence except for the amplitude decay when 62 272 272 L 1 t 7 AU 7 t st wra H W W H 11 397 N w H The attenuation constant controls how fast the wave decays Vzz M em cosat 82 envelope 2 A e V Vz Ale j 6 6 1 The wave fonNardtraveling wave is moving in the positive 2 direction Consider a lossless transmission line for simplicity or 0 V Z t A cosat 82 W Z t till v velocity Ff form ww 2m crest ofwave an 82 O The phase velocity VP is the velocity of a point on the wave such as the crest Set wt 82 constant Take the derivative with respect to time 8 dz 0 a dt Hence dz 0 d1 8 We thus have V Note this result holds also a 12 3 for a lossy line 1 Also we know that LC 2 lug C 2 d Hence Iossless line Let s now consider the backwardtraveling wave Iossless for simplicity V zt A cosat 82 1 11 v velocity 1 2 II V ZJ it sllllllif39llTl ji llillu Tl r Tl l lal Vz t lAl 6 0 cosat 82 VZ m 0 2010g10e Attenuation in dB dB 2 2010g10 lnx Use the following logarithm identity 10g10 x mm W az Therefore dB 2 20 20 mm mm Hence we have attenuation a dBm n Final attenuation formulas 0 dBm 2 attenuatlon In 10 attenuation 86859 a dBm Example coaxial cable copper conductors nonmagnetic y lL10 C L Fm In l a a 05 mm L 54112 Hm 323922mm 7239 a r 39 tanES 0001 cm 58gtlt107 Sm G 2 Sm 1n j f 500 MHz UHF a 2 R Qm 5 skin depth 0m5 27m 27219 qua m Dielectric conductivity is specified in terms of the loss tangent tan5 Q i cos cosogr a 05 mm b 32 mm 8r 22 tan6 0001 cm 58gtlt107Sm f 500 MHz UHF CI 05 mm b 32 mm 822 tan6 0001 7 58gtlt107Sm f 500 MHz UHF wLG ij c lij r quot quotf39f 1 17 faiLwr qu 1 3 HmH Alternative notations j a propagation constant k j7 ja propagation wavenumber Z Use the first I elegrapher equatior n 6 R1 Lg 62 at 1 2 J60 at a RI ijI 62 Next use Aeiyz I B eyz so a Z 7 Aeiyz B eyz 62 Hence we have 7Ae 86 2 RI ij1 Solving for the current I we have 1 R7jmj Ale 72 Be7z R ijG 103 AW ew Rij GUQ C Ae VZ Be Rij Define the complex Characteristic impedance Z0 12ij Zo Gch Then we have VZ Ae 86 12 2 21 Aeiyz B e72 ECE 3317 Prof David R Jackson Notes 11 Transmission Lines Standing Wave Ratio SWR and Generam zsd Re ection Coef cient Chapter 6 Consider a lossless transmission line that is terminated with a load sinusoidal source 392 Zg Iltzgt 39 7 Z0 VZ E9 ZL Z Z Z 0 ZO i FL L 0 C ZL Z0 VZ Aej z 11 e z 12 2 ZiAej Z FL e z 0 VZ Aej z1FL Hm 12 2 ZiAe W 1 FL em 0 VZ Aej z1FLejZ32 Denote FL 2 IFL l 61 Then we have VZ Adj32 1 F e L j 2 z The magnitude is VZ A 1 FL j 2 z 6 Maximum voltage VZ Vm iAi1 FL 2 z2 m max Maximum voltage Vz m 2 Km 2 IAKI I L 2 z 2 n mn0i1i2 F We then have 2 TL 1 S VSWR S OltD For the current we have Z 1 39 z 12Ae 15 Z o1iFLe 1 2396 j 1 IFL I ej 2 z 1 Iltzgt IAI Hence we have max 1z max ZIAILZL1IFLI 2 z 7r27m 0 min mIAIZilt1 Imgt 2 z27rm The current standing wave ratio is thus F L l FL 1 ISWR E Hence FL VzAej Z1FLejZ z VZll11 NM The voltage magnitude repeats in a distance Az 2 Az27r or 227 jAz 27 Vz Vz Note The complex voltage repeats every fl Mam1 1 39 2 z L 0 FL 6 I L ZL Z0 ZIFLIeW 1IFLI Vltzgt V V 1 W 139IFLI Ami2 Z 220 V Ae mz Special case of a real load impedance F ZLZORLZO L ZLZO RLZO 1 RL ZO 1FL RLZO 1 FL1 R Z SWR L O Z 1 L 0 Casea RLZZO SWR RLZ0RLZRL Z 1RL Zoj LZO R ZO Hence R Case b RL S Z0 1RLZ0 SWR RLZO RL ZO RL ZO 1 R 20 RLZORL ZO RLZ0 Hence WA R L Hence for a real load impedance we have SWR max amp 0 RL ZLRL 1 F JAM 2 1FZ where Note we go all the way around the crank diagram when 2 Changes by L 2 rL Z O24 j055 ZL 0 F 06 FL06e1 99 er V 1F 1O6 SWR ILI 40 V 1 FL1 O6 mm Vmax when 2IBZ 0 2739L39 Vmin when 2IBZ 739L39 3739L393 20 Z 199 rad Vmax V 1 FL16 Vmin V 1 rL04 Zmin 7z sz0921 23 227z VltZgt 05921 03421 00921 ET fem ui WE an I W W l Reverse problem Given What is the unknown load impedance 1IFL r 2 SWR 1FL 40 L SWR1 z z z z wu r 06e 99 1 2k 227z L 39 Defir Ie the geher elized reflectior I coefficier It at e poir It 20 Oh the lime Zg 3 53 1 1 e eihueoidel sour Ge Fl Zo Il ZL 20 Z O V ZO We Ger theh write Aeij zo eij15zizo I r 20 2j15zizo VZ Are 3z Zo 1quot Z0e z Zo Where A Ae j zo Zg 7 39V Z0 Hence Hence We then have ZZZ 20 311 ZO 1 1120 Calculate the reflection Coefficieht arid the ir Iput impedahce at ZL 7 Z0 1 ZL Z0 3 fL e 27 C 1 DCiO lzs v C 1 Deij z 0251 3 1rz lijB ZW 77 77 77 08 06 ECE 3317 Prof David R Jackson Notes 4 Maxwelll s Equations Adapted from notes by Prof Stuart A Long Four vector guantities E electric eld strength Voltmeter kgmsec3 Q electric ux density Coulmeterz Ampsecmz magnetic eld strength Ampmeter Ampm E magnetic ux density Webermeterz or Tesla kgAmp secz 4 each are functions of space and time 69 Exyzt 1 electric current density Ampmeterz pv electric charge density Coulmeter3 Ampsecm3 Sources generating electromagnetic fields Some common prefixes and the power of ten each represent a listed below femto f l 0 15 cehti C l 0 2 mega lI l 06 pico p 1012 deci d IO 1 gigs 3 109 haho r1 109 deka da 101 tera T 1012 micro 1 l 06 hecto h l 02 pets ID l O15 r r lilli r r l l 03 Kio K 1 03 timevarying differential form James Clerk Maxwell 1831 1879 James clerk Maxwell was a Seomsn mathemauman and heureuca physms t H s mustswgm cantacmevementwasthe dev m Ehemrumagneusm nas oeen eaueo the second onmoanon m pnysyos x a erthe no one named nut by saa NEWLEIH MaxWEH demuns trated that Ehectru ano magneu he d trave thruugh soaee m the farm ufvvaves ano aune unsia tspee ano magnene onenornena H s Wurk m of Ehemrumagneusm s cunswderad D be one of the greatesi advances m physms Wkipedia V X E Faraday s law Vx j Ja Q Ampere slaw at 0 Magnetic Gauss law V V ID ID UV Electric Gauss law lf39lquotlquot139lts 2 Tl l ljil ll39lllitf Vij 216 Q at WxgVJVa Qj at a OV1EVZD 220 Rate of decrease of electric charge per unit volume Flow of electric current out of volume 39 quot per unit volume apv at V i Integrate both sides over an arbitrary volume V apv lV J l at dV Apply the divergence theorem Physical interpretation S foul J aLdV pv dV This assumes thatthe V surface is stationary or TimeDependent Vsz a B x 8 D at at V Time ndependent Statics VgtltEQ V12pv VgtltH1 Vz30 Decouples E and gt E is a function of pV and is a function of i Characteristics of media relate QtoE and tog Free Space Q 50 E 50 permittivity 23924 22 12 0121 0 permeability 5 50 88541878x103912 Fm p 35 0 47rx10397 Hm exact exact value that is defined Free space in the phasor domain D 50 E 50 permittivity B 0 H 0 permeability This follows from the fact that aZt ltgt a where a is a real number In a material medium I 5 E 5 permittivity B it E permeability 8 80 8r gr relative permittivity I OILlr ur relative permittivity Depehdeht oh Ihdepehdeht of ihhomogeheous H or 8 Space frequehcy time field strehgth directior I of E homogehous rich dispersive staticFiery lir Iear isotropic dispersive hOh StatiOhar y hOh Iihear ahisotr opic Ecr a or u is a scalar quantity which means that E Q and in 5 a or u is a tensor can be written as a matrix Dx x x Dx 8x O 0 Ex Dy 2 yEy Dy 2 0 gy 0 Ey DZ 2 gZEZ DZ 0 0 82 E2 This results in E and Q being NOT proportional to each other ECE 3317 Prof David R Jackson Polarization of Wane waves I he polarizatioh of a plahe wave refers to the directior I of the electric field Vector lr I the time domeh S 272 ECZ t We assume that the Wave is travelrig lr I the positive Z directioh y Ey Phasor domain Assume Ex 2 a 2 real number 13 By be In general 8 phase of Ey phase of Ex Time Domain EU At Z 0 Ex 2 Re 616 a cos wt Ey Re 9615 em 9 cos wt Depending on b and three different cases arise Ex 61 Eyzib for0 for7z At 2 0 Ex 2 Reaej quot acoswt Ey Reib 61 2 ii cos wt y z X y b i shown for 0 I I X 1 a b 2 a 3 i AtZZO Ex 2 Reaej acoswt Ey Rebej ej Reaeij Zej acosati7r2 61 Sin wt 2 O 2 Ex2 Ey2 a2 0082 wta2 Slnz wt 2 Cl IEEE convention y 8 7z 2 RHCP f wt RHCP RHCP 39 Pk e w opposite rotation in space and time SzlExH 2 Ex Exza Hy772 Eyzijb H EyZ i jb x 77 77 This expression is actually valid for any plane wave that has both Ex and Ey Circular polarization is often used in communications to avoid problems with signal loss due to polarization changes gt Misalignment of receive antenna gt Reflections off of buildings gt Propagation through the atmosphere receive antenna The antenna will always receive a signal no matter how it is rotated about the z axis However for the same incident power density an optimum linearlypolarized wave will give the maximum output signal 3 dB higherthan from an incident CP wave Two ways in which circular polarization can be obtained 1 Use two identical antenna rotated by 90 and fed 90 out of phase antenna 2 antenna 1 This antenna will radiate a RHCP signal in the positive 2 direction and LHCP in the negative 2 direction antenna 2 antenna 1 signal feed line Z02Z012 power slitter Helical antenna for WLAN communication at 24 GHz httpenwikipedia orgwikiHeicaantenna An antenna that radiates circular polarization will also receive circular polarization of the same handedness and be blind to the opposite handedness antenna 2 antenna 1 RHCP It does not matter how the receive antenna is rotated about the z axis power combiner Includes all other cases that are not linear or circular 00r7z and i 7r2 or wzi 7r2 and b a EU I I I I 1 I I x I I I l I The tip of the electric field vector stays on an ellipse Here we give a proof that the tip of the electric field vector must stay on an ellipse I Ex 2 acos mt Ey bcosat bcoswtcos bsinwtsin SO 2 4qu a a or b b 2 Ey Excos sm bZ Ex2 Cl Squaring both sides we have b 2 b b 2 Ey2Ex2cos 2ExEycos jsmz b2 j Exz a b 2 b b 2 Ey2 Ex2 cos 8 2ExEy cos 8 sin2 8b2 j Exz a a Collecting terms we have 2 Ex2 cos2 sin2 By2 2ExEy cos 92 sin2 a a or 2 Ex2 ExEy 22008 3 Ey2 92 sin2 3 a a This is in the form of a quadratic expression AEXZ BExEy CEy2 D A B2 4AC determines the type of curve 4 cos2 3 4 a a ST cos2 B 1 b 2 so A 4 j sin2 lt0 Hence this is an ellipse Here we give a proof of the rotation property I Ex 2 acoswt Ey boosat bcoswtcos bsinwtsin EU E tan E y 2cos tanwtsin a X Take the derivative sec2 9 sec2ltwgtltwgtsm a 39 2 2 2 dt sm Kajcos sec a Olt lt7z g4 LHEP b 7 lt lt 0 g gt O RHEP proof complete method for deter mihihg the type of polerizetior n leads Ex We give a Simple ha Her e Ieft hahczleczl or right ded Im Ey 4g Ex Re I phasor dome y Z gt Ru Ie I he electric quotField Vector rotates from he lead i rig lagg ir IQ axie I time domaih Rule The electric field vector rotates from the leading axis to the lagging axis Im Ey phasor domain EU X I time domain Example E 1ji2 jejky What is this wave s polarization Example cont Eilji2jejky Im lEZ EZ leads Ex Re 1 Ex In time the wave rotates from the Z axis to the x axis Example cont E lji2je y gt LHEP or LHCP E x E Z 7 7r Note and 7 i so this is not LHCP Example cont E lji2je ky AAA quot 39 VV A x iLHjEP AR majoraxis minor axis AB CD gt1 Axial Ratio and Handedness sin 295 sin 27 sin 8 45 3 cf g 45 Tilt Angle AR COt tan 2239 tan 27 cos 8 95 gt 0 LHEP Eithegl iii39cf iambigm f lt 0 MEI ECE 3317 Prof David R Jackson Notes 20 ectianzguxlla Waveguides Chapter 5 Rectangular Waveguide Ty b 8u a cross section We assume that the boundary is a perfect electric conductor PEC We analyze the problem to solve for EZ or HZ all other fields come from these TMZ EZ only TEZ HZ only Z HZ 0 E2 i O kZE O Helmholtz equatior i PEG walls EZ 0 Oh bOUhdaI y Guided wave assumption EZ x y Z E20 xye 2 2 2 8x 32 52 aZEZ aZEZ 6x 2 kaZj kZEZ 322 2E 2E a 226 22 k2 k22Ez O 6x 6y Define 2 2 We then have 6 E2 6 E2 k2E 0 Note that kc isan I 6x2 632 C Z unknown at this pomt Dividing by the eXpjkzz term we have We solve the above equation by using the method of separation of variables We assume E20 xy XX YO where E20 xy Xquot YO Hence XquotYXYquot kchY Xquot Yquot 2 DIVIde beY kc X Y Hence X k02 Y Y This has the form F x Gy Both sides of the equation must be a constant Y 2 k3 constant X Xquot kf constant Set General solution Xx A Sinkxx B COSkxx Boundary conditions X0 0 1 Xa 0 2 1 g B 0 Xx Asinkxx 2 sinkxa 0 The second boundary condition is sinkxa o This gives us the following result kxazm m12 m7 9k x a Hence Xx Asinm x 61 Now we turn our attention to the Yy function General solution Yy Csinkyy D COSkyy Y y C sinky y D cosky y Boundary conditions Y0 0 3 W 0 4 3 D0 4 sinkyb0 Equation 4 gives us the following result kyb mz399 n 19 2 mt kyb Hence Y y C sin 9 Therefore we have E20 x y 2 X x my 2 flashile sin Cl New notation EzoxyAm The TMZ field inside the waveguide thus has the following form 2 2 2 Recall that ky kc kx Hence k k k Therefore the solution for kc is given by Summary of TM Solution for mn Mode HZ O 1x9WZZn nm xjnnyjev ma Cl mLH nLAm Note The number kc is the value of k for which the wavenumber k2 is zero where Set kgm 0 k kW M a 8 k m 27rf u8 k m Hence 27239f 8 m n 2 2 which gives us 275ny5 a b The cutoff frequency fc of the TMWI mode is then fmm 2 C 27 ya a b This may be written as fTMm Cd Cd 2 L C 27 a b V lug E E2 0 We now start with 6sz 6sz 6x2 6y2 k2 kHZO Using the separation of variables method again we have H20 w X x Yv where X x A sinkxx B coskxx Y y C sinky y D cosky y and k3 k kj k3 k2 k Boundary conditions Ty Exx0 0 Ey0y O Exxb0 Eyltaygto b 9 a cross section The result is HZ0 x y 2 Am cos 1mm cos a This can be shown by using the following equations aP12 a 32 6H2 Ex 2 2 2 2 q 0 k k2 8y k k2 6x 8y E 2 Jam 5H2 sz 5E2 Q 6H2 7 y k2 k ax k2 kj ay 5X Summary of TEZ Solution for mn Mode E2 O Hz x Z COS m7rx r r quot12quot cos JZ 6 Z a b Same formula for cutoff frequency as the TEZ case mQLZH nQL2m mm 72 00 Note the 00 TEZ mode is not valid since it violates the magnetic Gauss law HZxyzAOOeij VHxyz 0 IE2 x y Z IHZ x 322 myz x 7272 7 msrn A n 8111 8111 y e JZZ Z a b m7z x 7272 mn A n COS COS y e sz Z a b m n 2 my 2 747 2 kz k 7 7 I Same formula for both cases a b C I m7 2 n7z 2 mn d I I 39M m if 1 2 Z n 1 2 o 1 7 2 Same formula for both cases m 7 01 2 Gshsr sl formula for the wsvshumbsr 2 2 TMZ or TEZ mode kz k2 7 k3 with kc Note The 77171 hotstior I is suppressed her s k2 k1 7 kc k2 Recall k ug Wheh we are below cutoff it is COhvehieht to write Hehce k2 7Jkczi kz ijkc17kkc2 ijch17CQQ c2 7 ijkC1ifL72 Z l kch rffzl If xCWTHD2 32 I Recal that k2 Ja Hencewehave k1 Jf2 fgt akc1 fJ2 flt I C quotZ Dz C 2 32 I General behavior of the wavenumber k1fCf2 fgt fccd akc 1 f222 fltJZ 2 The quotdominantquot mode is the one with the lowest cutoff frequency Ty 0 m72 mzz d fa zztaHb b M a cross section Lowest TMZ mode TM11 Lowest TEZ mode TE1O TE l TMZ mzoplyza39 mszu nQLZm The dominant mode is the TE1O mode Formulas for the dominant TE1O mode IHZ DC 2 A10 003 eijkzz a k 1 fCfr2 fgt 056 1ffc2 flt What is the mode with the next highest cutoff frequency Cd E2 E2 midrib The next highest is the TE20 mode f20 2f10 useful operating region lt gt i l TE10 TE20 f6 Fields of the dominant TE10 mode Hz 36 y Z A10 0036 1792 61 Find the other fields from these equations E jau aZ ij aZ x 2 2 2 2 k kZ y k k 6x E jaw 5Hz sz y kZ kZZ 6x kZ kZZ 6y jwe 6 Z ij 6HZ Hx 2 2 1 2 2 k kZ 6y k kZ 6x ja8 6 Z ij 6 Z Hy 2 2 2 2 k kZ 6x k kZ y From these we find the other fields to be 1xayaz JA Elo Sin 6 1722 a A k Hx9yaz Z jEIO sln je szz my a jam 7 where E10 2 k2 k2 Zj yT E H gt x


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All subscriptions to StudySoup are paid in full at the time of subscribing. To change your credit card information or to cancel your subscription, go to "Edit Settings". All credit card information will be available there. If you should decide to cancel your subscription, it will continue to be valid until the next payment period, as all payments for the current period were made in advance. For special circumstances, please email


StudySoup has more than 1 million course-specific study resources to help students study smarter. If you’re having trouble finding what you’re looking for, our customer support team can help you find what you need! Feel free to contact them here:

Recurring Subscriptions: If you have canceled your recurring subscription on the day of renewal and have not downloaded any documents, you may request a refund by submitting an email to

Satisfaction Guarantee: If you’re not satisfied with your subscription, you can contact us for further help. Contact must be made within 3 business days of your subscription purchase and your refund request will be subject for review.

Please Note: Refunds can never be provided more than 30 days after the initial purchase date regardless of your activity on the site.